Please help me??? 3. For the gas system power illustrated in Figure 3 (with the air-standard analysis assumptions), calculate thermalefficiency, back work ratio, and Weycle-10P₁ = 100 kPaT₁ = 350 KOiaHeat exchangerHeat exchangerQoutT₁ - 1500 KTurbineWycle1T₁ - 1500 Kp= 1000 kPap= 100 kPaT₁ = 350 KFigure 3. Operating condition of a gas power system cycle (The P-V and T-S diagram)

Given :T1= 350 KT3= 1500 Kp1= 100 kPap2= 1000 kPaProcess 1-2 is isentropic compression so we need to…

Answered: 3. For the gas system power illustrated…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Please help me???

3. For the gas system power illustrated in Figure 3 (with the air-standard analysis assumptions), calculate thermal
efficiency, back work ratio, and Weycle
-10
P₁ = 100 kPa
T₁ = 350 K
Oia
Heat exchanger
Heat exchanger
Qout
T₁ - 1500 K
Turbine
Wycle
1
T₁ - 1500 K
p= 1000 kPa
p= 100 kPa
T₁ = 350 K
Figure 3. Operating condition of a gas power system cycle (The P-V and T-S diagram)

Expert Solution

Step 1: Given Data:

Given :

T₁= 350 K

T₃= 1500 K

p₁= 100 kPa

p₂= 1000 kPa

Process 1-2 is isentropic compression so we need to find T₂ by using $T_{2}$

$T subscript 2 over T subscript 1 equals open parentheses p subscript 1 over p subscript 2 close parentheses to the power of fraction numerator gamma minus 1 over denominator gamma end fraction end exponent T subscript 2 over 350 equals open parentheses 100 over 1000 close parentheses to the power of fraction numerator 1.4 minus 1 over denominator 1.4 end fraction end exponent T subscript 2 over 350 equals open parentheses 0.1 close parentheses to the power of 0.285 end exponent T subscript 2 equals 181.58 space K$

Similarly, process 3-4 is an isentropic expansion process, T₄ by using:

$T subscript 3 over T subscript 4 equals open parentheses P subscript 3 over P subscript 4 close parentheses to the power of fraction numerator gamma minus 1 over denominator gamma end fraction end exponent 1500 over T subscript 4 equals open parentheses 0.1 close parentheses to the power of 0.258 end exponent T subscript 4 equals 2891.28 space K$

Thermal Efficiency of the cycle is the fraction of net work done by cycle to the heat supplied in the cycle.

$eta equals fraction numerator N e t space W o r k space d o n e over denominator H e a t space a d d e d end fraction cross times 100 space space space equals fraction numerator negative 1228.9743 over denominator 1325.0121 end fraction cross times 100 space space space equals 92.75 percent sign$

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